If a, b and c are in AP and also in GP, then A) a = b ≠ c B) a ≠

1.	If a, b and c are in AP and also in GP, then A) a = b ≠ c B) a ≠ b = c C) a ≠ b ≠ c D) a = b = c
Answer» Correct option is (D) a = b = c Given that a, b, c are in AP and also in G.P. \(\therefore a+c=2b\) ______________(1) and \(ac=b^2\) ______________(2) \(\Rightarrow ac=(\frac{a+c}2)^2\) (From (1)) \(\Rightarrow ac=\frac{a^2+c^2+2ac}4\) \(\Rightarrow a^2+c^2+2ac=4ac\) \(\Rightarrow a^2+c^2-2ac=0\) \(\Rightarrow(a-c)^2=0\) \(\Rightarrow a-c=0\) \(\Rightarrow a=c\) Put a = c in equation (1), we get a+a = 2b \(\Rightarrow2b=2a\) \(\Rightarrow b=a\) Hence, a = b = c Correct option is D) a = b = c

If a, b and c are in AP and also in GP, then A) a = b ≠ c B) a ≠ b = c C) a ≠ b ≠ c D) a = b = c

Answer»

Correct option is (D) a = b = c

Given that a, b, c are in AP and also in G.P.

\(\therefore a+c=2b\) ______________(1)

and \(ac=b^2\) ______________(2)

\(\Rightarrow ac=(\frac{a+c}2)^2\) (From (1))

\(\Rightarrow ac=\frac{a^2+c^2+2ac}4\)

\(\Rightarrow a^2+c^2+2ac=4ac\)

\(\Rightarrow a^2+c^2-2ac=0\)

\(\Rightarrow(a-c)^2=0\)

\(\Rightarrow a-c=0\)

\(\Rightarrow a=c\)

Put a = c in equation (1), we get

a+a = 2b

\(\Rightarrow2b=2a\)

\(\Rightarrow b=a\)

Hence, a = b = c

Correct option is D) a = b = c